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Unformatted text preview: sses move along a (θ + α)so = speed of each + α),˙ (which you could also
circle,
the −
dt2 2
3R
2
obtain by calculating x + y ). The Lagrangian is therefore
˙
˙
√
d2
2g
2)
(θ − αα2 + mgR cos(θ −(α) + cos(θ + α)
=−
θ − α).
(348)
L = 2(m/2)R ˙
dt2
R
= mR2 α2 + 2mgR cos θ cos α.
˙
The normal coordinates are therefore (341) √ The equation of motion is
θ + α = A1 cos(ω1 t + φ1 ), 2g
where
ω1 =
,
3R
g cos θ
g cos θ
2
2mR α = −2mgR cos θ sin α =⇒ α ≈ −
¨
¨
α =⇒ √2g
ω=
. (342)
R
R
θ − α = A2 cos(ω2 t + φ2 ),
where
ω2 =
.
(349)
R
◦
Note that if θ = 0, then ω = g /R, as it should. And if θ = 90 , then ω = 0, which
Adding and subtracting these yields the angles, which when written in vector notamakes sense.
tion are
6.44. Oscillating hoop with a pendulum
θ
1
1
= B1
cos(ω2 t + φ2 ),
(350)
The positions of the masses cos(ω1 t + φ1 ) + B2
are
α
1
−1
◦
◦
(x, y ’s = R − A’s. If B θ = 0, we have θ)
(343)
where the B)1 are h...
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This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.
 Fall '13
 Brandenberger
 mechanics, Centrifugal Force, Force

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